What would you do if matrix A didn't begin with 1? How would you transform it to the identity matrix?

Assuming it doesn't begin with 0, divide every value in the augmented matrix's first row by the first number to get 1. Then, without actually changing the first row, take the first row multiplied by the negative of the first value of a row below it, and add it to that row, so that every value under the first 1 is a zero. Repeat this process for the second element in the second row, the third in the third row, and so on. When you get to the last element of the last row, start this process backwards so that the values above become 0s until you have the identity matrix.

What does he say at 1:11 ? Row etch-a-long form? What is "row etch-a-long form"? Thanks,

He says 'row echelon form.' Put any lines of all zeros on the bottom of the matrix, make the first entry in each row a one, and each entry in each row to the right of the first entry in the row above it. In form it is easy to see the solution to the equation.

Is it even possible to find inverse for a 5x5 matrix by the ref?

A bit late, but sure, why wouldn't it be possible?

What do you do if the matrix starts with 0?

If the top-left entry of the matrix is 0, you swap the top row with the bottom row (or whichever row is lowest down and has a non-zero first entry).

What if the the matrix begin with 5

If you mean that the top-left-most entry is a 5, then just divide that entire row by 5. Now the entry there is a 1, and you may continue as in the video. This technique generalizes even to situations when the matrix doesn't begin with 5. And you may need to repeat it later in the reduction as well. Hope this helps.

Determining the equation of a line perpendicular to another

Sal talks about perpendicular subspaces when he talks about the orthogonal complement at https://www.khanacademy.org/math/linear-algebra/alternate_bases/othogonal_complements/v/linear-algebra-orthogonal-complements.

You say in the beginning of the video that we know this matrix has the identy matrix when we get A to rref. Do we always have to check if a matrix has the Identity Matrix in the rref before we find the Inverse Matrix?

nevermind, You'll of course find out in the process...

Is this the only way to reduce to find the inverse. I am not talk about a different method but I mean is there only on possible inverse for each Matrix if there is an inverse.

inverse matrices are unique, provided it's invertible to begin with

Main content

Course: Linear algebra > Unit 2

Lesson 5: Finding inverses and determinants

Example of finding matrix inverse

Name: Example of finding matrix inverse
Uploaded: 2011-02-20T16:46:46Z
Description: Example of calculating the inverse of a matrix

Google Classroom

Example of calculating the inverse of a matrix. Created by Sal Khan.

Want to join the conversation?

Sort by:

Bushra Ibrahim
Posted 9 years ago. Direct link to Bushra Ibrahim's post “how we can fined the inve...”
how we can fined the inverse of a matrices of LU decomposition?
Button navigates to signup pageComment on Bushra Ibrahim's post “how we can fined the inve...”
(21 votes)
Answer
karbusplane
Posted 11 years ago. Direct link to karbusplane's post “What would you do if matr...”
What would you do if matrix A didn't begin with 1? How would you transform it to the identity matrix?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- wmessino
  Posted 11 years ago. Direct link to wmessino's post “Assuming it doesn't begin...”
  Assuming it doesn't begin with 0, divide every value in the augmented matrix's first row by the first number to get 1. Then, without actually changing the first row, take the first row multiplied by the negative of the first value of a row below it, and add it to that row, so that every value under the first 1 is a zero. Repeat this process for the second element in the second row, the third in the third row, and so on. When you get to the last element of the last row, start this process backwards so that the values above become 0s until you have the identity matrix.
  Button navigates to signup page
  (8 votes)
David
Posted 10 years ago. Direct link to David's post “What does he say at 1:11 ...”
What does he say at
1:11
? Row etch-a-long form? What is "row etch-a-long form"?

Thanks,
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Eric Anderson
  Posted 10 years ago. Direct link to Eric Anderson's post “He says 'row echelon form...”
  He says 'row echelon form.' Put any lines of all zeros on the bottom of the matrix, make the first entry in each row a one, and each entry in each row to the right of the first entry in the row above it. In form it is easy to see the solution to the equation.
  Comment on Eric Anderson's post “He says 'row echelon form...”
  (7 votes)
Niks Jansons
Posted 11 years ago. Direct link to Niks Jansons's post “Is it even possible to fi...”
Is it even possible to find inverse for a 5x5 matrix by the ref?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- michael.relleum
  Posted 2 years ago. Direct link to michael.relleum's post “A bit late, but sure, why...”
  A bit late, but sure, why wouldn't it be possible?
  Button navigates to signup page
  (3 votes)
Vamshi Aruru
Posted 8 years ago. Direct link to Vamshi Aruru's post “What do you do if the mat...”
What do you do if the matrix starts with 0?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Kyler Kathan
  Posted 8 years ago. Direct link to Kyler Kathan's post “If the top-left entry of ...”
  If the top-left entry of the matrix is 0, you swap the top row with the bottom row (or whichever row is lowest down and has a non-zero first entry).
  Button navigates to signup page
  (2 votes)
kbernard1405
Posted 8 years ago. Direct link to kbernard1405's post “What if the the matrix be...”
What if the the matrix begin with 5
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Robert
  Posted 8 years ago. Direct link to Robert's post “If you mean that the top-...”
  If you mean that the top-left-most entry is a 5, then just divide that entire row by 5. Now the entry there is a 1, and you may continue as in the video. This technique generalizes even to situations when the matrix doesn't begin with 5. And you may need to repeat it later in the reduction as well. Hope this helps.
  Button navigates to signup page
  (2 votes)
Björn Ívar Björnsson
Posted 13 years ago. Direct link to Björn Ívar Björnsson's post “You say in the beginning ...”
You say in the beginning of the video that we know this matrix has the identy matrix when we get A to rref. Do we always have to check if a matrix has the Identity Matrix in the rref before we find the Inverse Matrix?
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- Björn Ívar Björnsson
  Posted 13 years ago. Direct link to Björn Ívar Björnsson's post “nevermind, You'll of cour...”
  nevermind, You'll of course find out in the process...
  Button navigates to signup page
  (10 votes)
joshua.pate19
Posted 10 years ago. Direct link to joshua.pate19's post “Is this the only way to r...”
Is this the only way to reduce to find the inverse. I am not talk about a different method but I mean is there only on possible inverse for each Matrix if there is an inverse.
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- Bob Fred
  Posted 10 years ago. Direct link to Bob Fred's post “inverse matrices are uniq...”
  inverse matrices are unique, provided it's invertible to begin with
  Button navigates to signup page
  (4 votes)
Erick Soi
Posted 9 years ago. Direct link to Erick Soi's post “Determining the equation ...”
Determining the equation of a line perpendicular to another
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Tejas
  Posted 9 years ago. Direct link to Tejas's post “Sal talks about perpendic...”
  Sal talks about perpendicular subspaces when he talks about the orthogonal complement at https://www.khanacademy.org/math/linear-algebra/alternate_bases/othogonal_complements/v/linear-algebra-orthogonal-complements.
  Button navigates to signup page
  (2 votes)
Poke Dude
Posted 9 years ago. Direct link to Poke Dude's post “I keep on messing up my i...”
I keep on messing up my inverses, and the only change I do from other ways is I use R1 - R2, rather than -1 X R1 + R2. Is there a problem with that?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “They aren't the same, but...”
  They aren't the same, but both can be used to replace R1, or R2, if that helps reduce the matrix.
  Button navigates to signup page
  (1 vote)

Video transcript

In the last video, we stumbled upon a way to figure out the inverse for an invertible matrix. So, let's actually use that method in this video right here. I'm going to use the same matrix that we started off with in the last video. It seems like a fairly good matrix. We know that it's reduced row echelon form is the identity matrix, so we know it's invertable. So, let's find its inverse. The technique is pretty straightforward. You literally just apply the same transformations you would apply to this guy to get you to the identity matrix, and you would apply those same transformations to the identity matrix. That's because the collection of those transformations, if you represent them as matrixes, are really just the inverse of this guy. Let's just do it. So I'll create an augmented matrix here. Maybe I'll do it right here. Let me make it a little bit neater. First, I'll write a. It's 1, minus 1, 1. And then minus 1, 2, 1. Minus 1, 3, 4. And then I'll augment it with the identity matrix, with 1, 0, 0, 0, 1, 0, 0, 0, 1. Now, if I want to get a into reduced row echelon form, maybe I'll replace the second row. I'll keep the first row the same for now. Let me just draw it like this. The entire first row: 1, minus 1, minus 1. It's going to be augmented with 1, 0, 0. Keep the whole first row the same. Let's replace the second row with the second row plus the first row. Minus 1 plus 1 is 0. 2 plus minus 1 is 1. 3 plus minus 1 is 2. 0 plus 1 is 0. 1 plus-- oh, sorry. That was a tricky one. 0 plus 1 is 1. 1 plus 0 is 1. 0 plus 0 is 0. All I did is I added these two rows up. Now, this third row. Let me replace-- I want to get a zero here. Let me replace the third row with the third row minus the first row. 1 minus 1 is 0. 1 minus minus 1 is 2. 4 minus minus 1 is 5. 0 minus 1 is minus 1. 0 minus 0 is 0. And then 1 minus 0 is 1. Just like that. Now, what do we want to do? Well, we've gotten this far. We want to zero out that guy and that guy. Let's keep our second row the same. Let me write it down here. It's 0, 1, 2, and then you augmented it with 1, 1, 0. Just like that. And let's replace my first row with the first row plus the second row. 1 plus 0 is 1. Minus 1 plus 1 is 0. That's why I did that, to get a zero there. Minus 1 plus 2 is 1. 1 plus 1 is 2. 0 plus 1 is 1. 0 plus 0 is 0. And now, I also want to zero out this guy right here. Let's replace the third row with the third row minus 2 times the second row. 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 5 minus 2 times 2 is 5 minus 4, that's 1. Minus 1 minus 2 times 1-- that's minus 1 minus 2-- is minus 3. 0 minus 2 times 1, that's minus 2. And then, 1 minus 2 times 0 is just 1 again. All right, home stretch. Now, I just want to zero out these guys right here. All right, so just let me keep my third row the same. Let me switch colors, keep things colorful. It's 0, 0, 1. We're going to augment it with minus 3, minus 2, and 1. Now, let's replace our first row with the first row minus the third row. 1 minus 0 is 1. 0 minus 0 is 0. 1 minus 1 is 0. 2 minus minus 3, that's 5. 1 minus minus 2 is 3. 0 minus 1 is minus 1. Now, let's replace the second row with the second row minus 2 times the third row. 0 minus 2 times 0 is 0. 1 minus 2 times 0 is 0. 2 minus 2 times 1 is-- I'm sorry, I just-- oh, whoops. Let me-- we have to be very careful not to make any careless mistakes. 0 minus 2 times 0 is 0. 1 minus 2 times 0 is 1. It's not 0. 2 minus 2 times 1 is 0. 1 minus 2 times minus 3-- that is 1 plus 2 times 3-- that is 7. 1 minus 2 times minus 2, that's 1 plus 4, which is 5. And then, 0 minus 2 times 1, so that's minus 2. And just like that, we've gotten the A part of our augmented matrix into reduced row echelon form. This is the reduced row echelon form of A. And when you apply those exact same transformations-- because if you think about it, that series of matrix products that got you from this to the identity matrix-- that, by definition, is the identity matrix. So you apply those same transformations to the identity matrix, you're going to get the inverse of A. This right here is A inverse. And we have solved for the inverse, and it actually wasn't too painful.